Question:

Show that a.(b×c) is equal in magnitude to the volume of the parallelopiped formed on the three vectors, a, b and c.

Answer:

Step 1: Recall the definition of the dot product: a.b = |a||b|cosθ, where θ is the angle between a and b.

Step 2: Recall the definition of the cross product: b×c = |b||c|sinθ, where θ is the angle between b and c.

Step 3: Substitute the definitions of the dot product and cross product into the equation a.(b×c) to get: a.(b×c) = |a||b||c|cosθsinθ.

Step 4: Recall the formula for the volume of a parallelopiped: V = |a||b||c|sinθ, where θ is the angle between a and b.

Step 5: Substitute the formula for the volume of a parallelopiped into the equation a.(b×c) = |a||b||c|cosθsinθ to get: a.(b×c) = V.

Therefore, a.(b×c) is equal in magnitude to the volume of the parallelopiped formed on the three vectors, a, b and c.

Question:

A car weighs 1800 kg. The distance between its front and back axles is 1.8 m. Its centre of gravity is 1.05 m behind the front axle. Determine the force exerted by the level ground on each front wheel and each back wheel.

Answer:

1. Calculate the weight of the car (1800 kg).

2. Calculate the distance between the front and back axles (1.8 m).

3. Calculate the centre of gravity (1.05 m behind the front axle).

4. Calculate the force exerted by the level ground on each front wheel: F = (1800 kg x 9.8 m/s2) x (1.05 m / 1.8 m) = 1023.33 N

5. Calculate the force exerted by the level ground on each back wheel: F = (1800 kg x 9.8 m/s2) x (0.75 m / 1.8 m) = 759 N

Question:

To maintain a rotor at a uniform angular speed of 200 rad s^−1, an engine needs to transmit a torque of 180 N m. What is the power required by the engine ? (Note: uniform angular velocity in the absence of friction implies zero torque. In practice, applied torque is needed to counter frictional torque). Assume that the engine is 100% efficient.

Answer:

Step 1: Calculate the angular velocity (ω) of the rotor: ω = 200 rad s^−1

Step 2: Calculate the torque (T) applied by the engine: T = 180 N m

Step 3: Calculate the power (P) required by the engine: P = T x ω P = (180 N m) x (200 rad s^−1) P = 36,000 W

Step 4: Calculate the power (P) required by the engine with 100% efficiency: P = 36,000 W P (100% efficiency) = 36,000 W

Question:

A hoop of radius 2 m weighs 100 kg. It rolls along a horizontal floor so that its centre of mass has a speed of 20 cm/s. How much work has to be done to stop it?

Answer:

1. Calculate the moment of inertia of the hoop: I = mr2 = 100 x (2)2 = 400 kg m2

2. Calculate the kinetic energy of the hoop: KE = ½Iω2 = ½ x 400 x (20/100)2 = 0.8 J

3. Calculate the work done to stop the hoop: W = KE = 0.8 J

Question:

Two discs of moments of inertia I1​ and I2​ about their respective axes (normal to the disc and passing through the centre), and rotating with angular speeds ω1​ and ω2​ are brought into contact face to face with their axes of rotation coincident. (a) What is the angular speed of the two-disc system? (b) Show that the kinetic energy of the combined system is less than the sum of the initial kinetic energies of the two discs. How do you account for this loss in energy? Take ω1​=ω2​.

Answer:

(a) The angular speed of the two-disc system is equal to the average of the angular speeds of the two discs, i.e. ω = (ω1 + ω2)/2.

(b) The kinetic energy of the combined system is given by:

K = (I1 + I2)ω2/2

The initial kinetic energies of the two discs are given by:

K1 = I1ω12/2 K2 = I2ω22/2

The sum of the initial kinetic energies is:

K1 + K2 = (I1ω12 + I2ω22)/2

Since ω1 = ω2, the sum of the initial kinetic energies is equal to:

K1 + K2 = (I1 + I2)ω2/2

This is the same as the kinetic energy of the combined system. Therefore, the kinetic energy of the combined system is equal to the sum of the initial kinetic energies of the two discs.

The loss in energy is due to the friction between the two discs as they come into contact. This friction causes some of the energy to be converted into heat, which is then dissipated.

Question:

Read each statement below carefully, and state, with reasons, if it is true or false; (a) During rolling, the force of friction acts in the same direction as the direction of motion of the CM of the body. (b) The instantaneous speed of the point of contact during rolling is zero. (c) The instantaneous acceleration of the point of contact during rolling is zero. (d) For perfect rolling motion, work done against friction is zero. (e) A wheel moving down a perfectly frictionless inclined plane will undergo slipping (not rolling) motion

Answer:

(a) False. During rolling, the force of friction acts in the opposite direction as the direction of motion of the CM of the body.

(b) True. The instantaneous speed of the point of contact during rolling is zero.

(c) False. The instantaneous acceleration of the point of contact during rolling is not zero.

(d) True. For perfect rolling motion, work done against friction is zero.

(e) True. A wheel moving down a perfectly frictionless inclined plane will undergo slipping (not rolling) motion.

Question:

Show that the area of the triangle contained between the vectors a and b is one half of the magnitude of a×b.

Answer:

Step 1: Recall that the area of a triangle is equal to one half of the base multiplied by the height.

Step 2: Define a×b as the magnitude of the cross product of the two vectors.

Step 3: Use the cross product formula to calculate the magnitude of a×b.

Step 4: Calculate the height of the triangle contained between the vectors a and b.

Step 5: Multiply the base by the height of the triangle and divide the result by two to get the area.

Step 6: Show that the area of the triangle is equal to one half of the magnitude of a×b.

Question:

(a) Find the moment of inertia of a sphere about a tangent to the sphere, given the moment of inertia of the sphere about any of its diameters to be 2MR2^/5, where M is the mass of the sphere and R is the radius of the sphere. (b) Given the moment of inertia of a disc of mass M and radius R about any of its diameters to be MR^2/4, find its moment of inertia about an axis normal to the disc and passing through a point on its edge.

Answer:

(a) Moment of inertia of a sphere about a tangent to the sphere = 2MR^2/5

(b) Moment of inertia of a disc about an axis normal to the disc and passing through a point on its edge = MR^2/2

Question:

A solid cylinder of mass 20 kg rotates about its axis with angular speed 100 rad s−1. The radius of the cylinder is 0.25 m. What is the kinetic energy associated with the rotation of the cylinder? What is the magnitude of angular momentum of the cylinder about its axis?

Answer:

Kinetic Energy = 1/2 × mass × angular velocity² × radius²

Kinetic Energy = 1/2 × 20 kg × (100 rad s−1)² × (0.25 m)²

Kinetic Energy = 25,000 J

Angular Momentum = mass × angular velocity × radius

Angular Momentum = 20 kg × 100 rad s−1 × 0.25 m

Angular Momentum = 5,000 kg m² s−1

Question:

A solid sphere rolls down on two different inclined planes of the same heights but different angles of inclination. (a) Will it reach the bottom with the same speed in each case? (b) Will it take longer to roll down one plane than the other? (c) If so, which one and why?

Answer:

a) No, it will not reach the bottom with the same speed in each case.

b) Yes, it will take longer to roll down one plane than the other.

c) The plane with the steeper angle of inclination will take longer to roll down, because it has a greater force of gravity acting on it, making it harder for the sphere to roll down.

Question:

Separation of Motion of a system of particles into motion of the centre of mass and motion about the centre of mass : (a) Show p=pi′​+mi​V where pi​ is the momentum of the ith particle (of mass mi​) and pi′​=mi​vi′​. Note vi′​ is the velocity of the ith particle relative to the centre of mass. Also, prove using the definition of the center of mass ∑pi′​=0 (b) Show K=K′+1​/2MV^2 where K is the total kinetic energy of the system of particles, K′ is the total kinetic energy of the system when the particle velocities are taken with respect to the centre of mass and MV^2/2 is the kinetic energy of the translation of the system as a whole (i.e. of the centre of mass motion of the system). The result has been used in Sec. 7.14. (c) Show L=L′+R×MV where L′=ri′​×pi′​ is the angular momentum of the system about the centre of mass with velocities taken relative to the centre of mass. Remember ri′​=ri​−R; rest of the notation is the standard notation used in the chapter. Note L′ and MR×V can be said to be angular momenta, respectively, about and of the centre of mass of the system of particles.(d) Show dL′​/dt=∑ri′​×dp′​/dt Further, show that dL′​/dt=τext′​ where τext′​ is the sum of all external torques acting on the system about the centre of mass.(Hint : Use the definition of centre of mass and Newtons Third Law. Assume the internal forces between any two particles act along the line joining the particles.)

Answer:

(a) p=pi′​+mi​V

Proof:

Let R be the centre of mass of the system of particles. Then,

R=∑mi​ri​/∑mi​

Let vi​ be the velocity of the ith particle. Then,

vi′​=vi​−V

where V is the velocity of the centre of mass.

Therefore,

pi′​=mi​vi′​=mi​(vi​−V)

Also,

pi​=mi​vi​

Therefore,

p=pi′​+mi​V

∑pi′​=0

Proof:

We know that

pi′​=mi​vi′​=mi​(vi​−V)

Therefore,

∑pi′​=∑mi​vi′​=∑mi​(vi​−V)

But,

V=∑mi​vi​/∑mi​

Therefore,

∑pi′​=∑mi​vi′​=∑mi​(vi​−∑mi​vi​/∑mi​)

=∑mi​vi​−∑mi​∑mi​vi​/∑mi​

=∑mi​vi​−∑mi​vi​

=0

(b) K=K′+1​/2MV^2

Proof:

We know that

K=∑mi​vi​^2/2

Also,

K′=∑mi​vi′​^2/2

Therefore,

K=K′+∑mi​vi′​^2/2−∑mi​vi′​^2/2

=K′+1​/2MV^2

where M is the total mass of the system.

(c) L=L′+R×MV

Proof:

We know that

L=∑ri​×pi​

Also,

L′=∑ri′​×pi′​

where

ri′​=ri​−R

and

pi′​=mi​vi′​

Therefore,

L=L′+∑ri′​×pi′​−∑ri′​×pi′​

=L′+R×MV

where M is the total mass of the system and V is the velocity of the centre of mass.

(d) dL′​/dt=∑ri′​×dp′​/dt

Proof:

We know that

dL′​/dt=∑ri′​×dp′​/dt

Also,

dp′​/dt=d(mi​vi′​)/dt

=mi​dvi′​/dt

Therefore,

dL′​/dt=∑mi​ri′​×dvi′​/dt

Further,

dL′​/dt=τext′​

Proof:

We know that

dL′​/dt=∑mi​ri′​×dvi′​/dt

Also,

τext′​=∑ri′​×Fext​

where Fext​ is the external force acting on the system.

Therefore,

dL′​/dt=∑mi​ri′​×dvi′​/dt=∑ri′​×Fext​=τext′​

where Fext​ is the external force acting on the system.

Note: This result can be used to prove Newtons Third Law.

Question:

Give the location of the centre of mass of a (i) sphere, (ii) cylinder, (iii) ring, and (iv) cube, each of uniform mass density. Does the centre of mass of a body necessarily lie inside the body ?

Answer:

(i) The centre of mass of a sphere is at its centre.

(ii) The centre of mass of a cylinder is at its centre.

(iii) The centre of mass of a ring is at its centre.

(iv) The centre of mass of a cube is at its centre.

No, the centre of mass of a body does not necessarily lie inside the body. For example, the centre of mass of a lopsided object such as a dumbbell would not lie inside the body.

Question:

A child sits stationary at one end of a long trolley moving uniformly with a speed V on a smooth horizontal floor. If the child gets up and runs about on the trolley in any manner, what is the speed of the CM of the (trolley + child) system ?

Answer:

1. The center of mass (CM) of the (trolley + child) system is the point where the total mass of the system is balanced.

2. The mass of the trolley is constant and the mass of the child is constant.

3. The speed of the trolley is V and the speed of the child is 0 when the child is sitting stationary.

4. When the child gets up and runs about on the trolley, the speed of the child increases.

5. The speed of the CM of the system is the average of the speed of the trolley and the speed of the child.

6. Therefore, the speed of the CM of the (trolley + child) system is the average of V and the speed of the child.

Question:

Two particles, each of mass m and speed v, travel in opposite directions along parallel lines separated by a distance d. Show that the vector angular momentum of the two particle system is the same whatever be the point about which the angular momentum is taken.

Answer:

1. Define the vector angular momentum of a particle as the cross product of its position vector and its linear momentum vector.

2. Express the position vector of each particle in terms of the distance d between them and the point about which the angular momentum is taken.

3. Calculate the linear momentum vector of each particle using its mass and speed.

4. Calculate the vector angular momentum of each particle by taking the cross product of its position vector and its linear momentum vector.

5. Show that the vector angular momentum of the two particle system is the same whatever be the point about which the angular momentum is taken by adding the angular momentum vectors of the two particles.

Question:

Prove the result that the velocity v of translation of a rolling body (like a ring, disc, cylinder or sphere) at the bottom of an inclined plane of a height h is given by v^2=2gh​/1+k^2​/R^2 using dynamical consideration (i.e. by consideration of forces and torques). Note k is the radius of gyration of the body about its symmetry axis, and R is the radius of the body. The body starts from rest at the top of the plane.

Answer:

1. Consider the body at the top of the inclined plane. At this point, the body is at rest and the only force acting on it is the force of gravity, which is directed downwards.

2. As the body begins to roll down the inclined plane, it experiences a force of friction, which acts parallel to the surface of the plane. This force of friction opposes the motion of the body and slows it down.

3. At the same time, the body experiences a torque due to the force of gravity. This torque causes the body to rotate about its symmetry axis.

4. As the body rolls down the inclined plane, it gains speed due to the force of gravity. This increase in speed is opposed by the force of friction, which acts to slow the body down.

5. At the bottom of the inclined plane, the body has reached its maximum speed. At this point, the force of friction is equal to the force of gravity, and the torque due to gravity is equal to the torque due to friction.

6. The velocity of the body at the bottom of the inclined plane can be calculated by equating the forces and torques acting on the body. The equation for the velocity is given by v^2=2gh/1+k^2/R^2, where h is the height of the inclined plane, k is the radius of gyration of the body about its symmetry axis, and R is the radius of the body.

Question:

Torques of equal magnitude are applied to a hollow cylinder and a solid sphere, both having the same mass and radius. The cylinder is free to rotate about its standard axis of symmetry, and the sphere is free to rotate about an axis passing through its centre. Which of the two will acquire a greater angular speed after a given time?

Answer:

1. Torques of equal magnitude are applied to a hollow cylinder and a solid sphere, both having the same mass and radius.

2. The cylinder is free to rotate about its standard axis of symmetry, and the sphere is free to rotate about an axis passing through its centre.

3. Angular acceleration is directly proportional to the applied torque and inversely proportional to the moment of inertia.

4. The moment of inertia of a solid sphere is greater than that of a hollow cylinder.

5. Therefore, the cylinder will acquire a greater angular speed after a given time.

Question:

A solid cylinder rolls up an inclined plane of angle of inclination 30o. At the bottom of the inclined plane the center of mass of the cylinder has a speed of 5m/s. (a) How far will the cylinder go up the plane? (b) How long will it take to return to the bottom?

Answer:

(a) To calculate the distance the cylinder will go up the plane, we must use the equation d = vtcos(theta), where d is the distance, v is the velocity, t is the time, and theta is the angle of inclination. Since we know the velocity and the angle of inclination, we can solve for the time by rearranging the equation to t = d/(v*cos(theta)). We can then plug in the values for the velocity and the angle of inclination to calculate the time it takes the cylinder to reach the top of the inclined plane.

(b) To calculate the time it takes for the cylinder to return to the bottom of the inclined plane, we can use the equation t = d/(v*sin(theta)), where d is the distance, v is the velocity, t is the time, and theta is the angle of inclination. We can then plug in the values for the velocity and the angle of inclination to calculate the time it takes the cylinder to return to the bottom of the inclined plane.

Question:

A man stands on a rotating platform, with his arms stretched horizontally holding a 5 kg weight in each hand. The angular speed of the platform is 30 revolutions per minute. The man then brings his arms close to his body with the distance of each weight from the axis changing from 90cm to 20cm. The moment of inertia of the man together with the platform may be taken to be constant and equal to 7.6 kg m^2. (a) What is his new angular speed? (Neglect friction.) (b) Is kinetic energy conserved in the process? If not, from where does the change come about?

Answer:

a) The new angular speed of the man and the platform can be calculated using the equation of conservation of angular momentum.

Angular momentum = Iω

Where I is the moment of inertia and ω is the angular speed.

Therefore,

I1ω1 = I2ω2

Where I1 and ω1 are the initial moment of inertia and angular speed, and I2 and ω2 are the final moment of inertia and angular speed.

I1 = 7.6 kg m^2

ω1 = 30 revolutions per minute

I2 = 7.6 kg m^2 (as the moment of inertia is constant)

ω2 =?

Substituting the values in the equation,

7.6 kg m^2 (30 revolutions per minute) = 7.6 kg m^2 ω2

ω2 = 30 revolutions per minute

Therefore, the new angular speed of the man and the platform is 30 revolutions per minute.

b) No, kinetic energy is not conserved in the process. The change in kinetic energy comes from the change in the distance of the weights from the axis of rotation. As the distance of the weights from the axis of rotation decreases from 90 cm to 20 cm, the kinetic energy decreases.

Question:

(a) Prove the theorem of perpendicular axes. (Hint : Square of the distance of a point (x, y) in the xy plane from an axis through the origin and perpendicular to the plane is x^2+y^2). (b) Prove the theorem of parallel axes. (Hint : If the centre of mass is chosen to be the origin ∑mi​ri​=0 ).

Answer:

(a) Proof of Theorem of Perpendicular Axes:

Let P(x,y) be a point in the xy plane with distance d from the origin.

Let the line through the origin and perpendicular to the xy plane be the x’ axis.

Then, the distance of the point P from the x’ axis is d’.

By the Pythagorean theorem, we have d2 = d’2 + x2.

Therefore, d2 = d’2 + x2 + y2.

This proves that the square of the distance of a point (x,y) in the xy plane from an axis through the origin and perpendicular to the plane is x2 + y2.

(b) Proof of Theorem of Parallel Axes:

Let M1, M2, …, Mn be the masses of n particles in a plane. Let the centre of mass of these particles be the origin O.

Let the line through the origin and perpendicular to the plane be the x’ axis.

Let ri be the vector from the origin O to the particle Mi.

Then, the x’ coordinate of the particle Mi is xi = ri .cosθ, where θ is the angle between the vector ri and the x’ axis.

The moment of inertia I of the particles about the x’ axis is given by I = ∑Mi xi2.

Substituting for xi, we get I = ∑Mi (ri .cosθ)2.

Now, ri .cosθ = |ri|.cosθ = |ri| cosθ.

Therefore, I = ∑Mi |ri|2.cos2θ.

Since ∑Mi ri = 0, we have ∑Mi |ri| = 0.

Therefore, I = 0.

This proves that the moment of inertia of n particles in a plane about an axis through the origin and parallel to the plane is zero.

Question:

A solid disc and a ring, both of radius 10 cm are placed on a horizontal table simultaneously, with initial angular speed equal to 10π rad s^−1. Which of the two will start to roll earlier ? The co-efficient of kinetic friction is μk​=0.2.

Answer:

1. Determine the angular speed of the disc and the ring. The angular speed of the disc and the ring is 10π rad s^−1.

2. Calculate the angular acceleration of the disc and the ring. The angular acceleration of the disc and the ring is 0, since they have an initial angular speed of 10π rad s^−1.

3. Calculate the static and kinetic friction coefficients. The static friction coefficient is μs = 0.2 and the kinetic friction coefficient is μk = 0.2.

4. Determine which of the two will start to roll earlier. The disc will start to roll earlier since it has a larger moment of inertia than the ring. This means that the disc will require less torque to overcome the static friction and start to roll.

Question:

A solid cylinder rolls up an inclined plane of angle of inclination 30o. At the bottom of the inclined plane, the centre of mass of the cylinder has a speed of 5m/s. 1.How far will the cylinder go up the plane? 2.How long will it take to return to the bottom?

Answer:

1. The distance traveled by the cylinder up the inclined plane can be calculated using the equation: Distance = (Initial Velocity) x (Time). The initial velocity of the cylinder is 5m/s.

2. The time taken by the cylinder to return to the bottom of the inclined plane can be calculated using the equation: Time = (Distance) / (Initial Velocity). The distance traveled by the cylinder is the same as the distance traveled up the inclined plane. Therefore, the time taken by the cylinder to return to the bottom of the inclined plane is equal to the time taken to travel up the inclined plane.

Question:

Find the components along the x, y, z axes of the angular momentum l of a particle, whose position vector is r with components x, y, z and momentum is p with components px​,py​ and pz​. Show that if the particle moves only in the x-y plane the angular momentum has only a z-component.

Answer:

1. The angular momentum l of the particle is given by: l = r × p

2. Therefore, the components of the angular momentum l along the x, y, and z axes can be calculated as: lx = ypz - zpy ly = zpx - xpz lz = xpy - ypx

3. If the particle moves only in the x-y plane, then the z-component of the position vector r and the z-component of the momentum vector p will be zero.

4. Substituting this in the equations for lx, ly, and lz, we get: lx = 0 ly = 0 lz = xpy

5. Therefore, if the particle moves only in the x-y plane, the angular momentum has only a z-component.

Question:

(a) A child stands at the centre of a turntable with his two arms outstretched. The turntable is set rotating with an angular speed of 40 rev/min. How much is the angular speed of the child if he folds his hands back and thereby reduces his moment of inertia to 2/5 times the initial value ? Assume that the turntable rotates without friction. (b) Show that the childs new kinetic energy of rotation is more than the initial kinetic energy of rotation. How do you account for this increase in kinetic energy?

Answer:

(a) The angular speed of the child is 40 rev/min x (2/5) = 16 rev/min.

(b) The increase in kinetic energy is due to the decrease in moment of inertia. Since the angular speed is increased by a factor of 2/5, the kinetic energy is increased by a factor of (2/5)^2 = 4/25. Therefore, the new kinetic energy is 4/25 times the initial kinetic energy.

Question:

A rope of negligible mass is wound round a hollow cylinder of mass 3 kg and radius 40 cm. What is the angular acceleration of the cylinder if the rope is pulled with a force of 30 N? What is the linear acceleration of the rope Assume that there is no slipping.

Answer:

1. Calculate the moment of inertia of the cylinder: I = (1/2) mr^2 I = (1/2) (3 kg)(0.4 m)^2 I = 0.48 kg m^2

2. Calculate the angular acceleration of the cylinder: α = F/I α = (30 N)/(0.48 kg m^2) α = 62.5 rad/s^2

3. Calculate the linear acceleration of the rope: a = αr a = (62.5 rad/s^2)(0.4 m) a = 25 m/s^2

Question:

From a uniform disk of radius R, a circular hole of radius R/2 is cut out. The centre of the hole is at R/2 from the centre of the original disc.Locate the centre of gravity of the resulting flat body.

Answer:

1. Take the centre of the original disc as the origin.

2. Consider the two regions of the disc, the circular hole and the remaining part of the disc.

3. Calculate the area of the circular hole. The area of a circle is given by A = πr2, where r is the radius of the circle. Thus, the area of the circular hole is A = π(R/2)2 = πR2/4.

4. Calculate the area of the remaining part of the disc. The area of a circle is given by A = πr2, where r is the radius of the circle. Thus, the area of the remaining part of the disc is A = πR2.

5. Calculate the distance of the centre of gravity of the circular hole from the origin. The centre of gravity of a circle is located at its centre, which is at a distance of R/2 from the origin.

6. Calculate the distance of the centre of gravity of the remaining part of the disc from the origin. The centre of gravity of a circle is located at its centre, which is at a distance of R from the origin.

7. Calculate the total area of the disc. The total area of the disc is the sum of the area of the circular hole and the area of the remaining part of the disc. Thus, the total area of the disc is A = πR2 + πR2/4 = 5πR2/4.

8. Calculate the distance of the centre of gravity of the disc from the origin. The centre of gravity of a body is given by x = (A1x1 + A2x2 + … + Anxn)/(A1 + A2 + … + An), where A1, A2, …, An are the areas of the different parts of the body, and x1, x2, …, xn are the distances of the centres of gravity of the different parts of the body from the origin. Thus, the distance of the centre of gravity of the disc from the origin is x = (πR2/4 x R/2 + πR2 x R)/(5πR2/4) = 4R/5.

9. The centre of gravity of the resulting flat body is located at a distance of 4R/5 from the origin.

Question:

A metre stick is balanced on a knife edge at its centre. When two coins, each of mass 5 g are put one on top of the other at the 12.0 cm mark, the stick is found to be balanced at 45.0 cm. What is the mass of the metre stick?

Answer:

Step 1: Calculate the distance between the coins and the centre of the stick.

Distance = 12.0 cm - 45.0 cm = 33.0 cm

Step 2: Calculate the total mass of the coins.

Mass of coins = 5 g x 2 = 10 g

Step 3: Calculate the mass of the metre stick.

Mass of metre stick = 10 g x (33.0 cm/45.0 cm) = 7.3 g

Question:

In the HCl molecule, the separation between the nuclei of the two atoms is about 1.27 A˚(1A˚=10−10 m). Find the approximate location of the CM of the molecule, given that a chlorine atom is about 35.5 times as massive as a hydrogen atom and nearly all the mass of an atom is concentrated in its nucleus.

Answer:

Step 1: Calculate the mass of the hydrogen atom (mH) and the mass of the chlorine atom (mCl).

mH = 1.0079 g/mol mCl = 35.5 x mH = 35.45 g/mol

Step 2: Calculate the distance from the center of the hydrogen nucleus to the center of the chlorine nucleus (d).

d = 1.27 A˚ = 1.27 x 10-10 m

Step 3: Calculate the mass of the hydrogen nucleus (mHN) and the mass of the chlorine nucleus (mClN).

mHN = 1.0078 g/mol mClN = 35.45 g/mol

Step 4: Calculate the total mass of the HCl molecule (mT).

mT = mHN + mClN = 36.46 g/mol

Step 5: Calculate the location of the CM of the molecule (rCM).

rCM = (mHN x d) / mT rCM = (1.0078 x 1.27 x 10-10) / 36.46 rCM = 3.48 x 10-12 m

Question:

A bullet of mass 10 g and speed 500 m/s is fired into a door and gets embedded exactly at the center of the door. The door is 1.0 m wide and weighs 12 kg. It is hinged at one end and rotates about a vertical axis practically without friction. Find the angular speed of the door just after the bullet embeds into it.

Answer:

1. Determine the moment of inertia of the door: I = (1/3)m x L2 = (1/3) (12 kg) (1.0 m)2 = 4 kg-m2

2. Calculate the angular momentum of the bullet: L = m x v x r = (10 g) (500 m/s) (0.5 m) = 2500 kg-m/s

3. Calculate the angular speed of the door: ω = L/I = (2500 kg-m/s) / (4 kg-m2) = 625 rad/s

Question:

A cylinder of mass 10 kg and radius 15 cm is rolling perfectly on a plane of inclination 30o. The coefficient of static friction μs​=0.25. (a) How much is the force of friction acting on the cylinder ? (b) What is the work done against friction during rolling ? (c) If the inclination θ of the plane is increased, at what value of θ does the cylinder begin to skid, and not roll perfectly ?

Answer:

(a) The force of friction acting on the cylinder is given by the equation F = μsmg, where μs is the coefficient of static friction, m is the mass of the cylinder and g is the acceleration due to gravity. Therefore, the force of friction acting on the cylinder is F = 0.2510*9.81 = 24.525 N.

(b) The work done against friction during rolling is given by the equation W = Fd, where F is the force of friction and d is the distance travelled. Since the cylinder is rolling perfectly, the distance travelled is equal to the circumference of the cylinder, which is 2πr, where r is the radius of the cylinder. Therefore, the work done against friction during rolling is W = 24.5252π*0.15 = 37.7 J.

(c) The coefficient of static friction is a measure of the maximum frictional force that can be applied to an object before it begins to skid, and not roll perfectly. Therefore, the cylinder will begin to skid, and not roll perfectly, when the angle of inclination θ exceeds the value of the coefficient of static friction, which is 0.25 in this case. Therefore, the cylinder will begin to skid, and not roll perfectly, when the angle of inclination θ exceeds 0.25.